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What is the rightmost non-zero digit of 20!?

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Re: PS - Rightmost nonzero digit & product of factors [#permalink] New post 24 Sep 2010, 06:27
Consider the seven numbers ending in 1,3,4,6,7,8,9
Last digit can be calculated as :
1 & 3 --> 3
3 & 4 --> 2
2 & 6 --> 2
2 & 7 --> 4
4 & 8 --> 2
2 & 9 --> 8

And in set 4 there are exactly 2 such subsets of seven numbers

So last digit is that of 8 & 8 --> 4
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Re: PS - Rightmost nonzero digit & product of factors [#permalink] New post 08 May 2013, 02:44
sergbov123 wrote:
Q1) What is the rightmost non-zero digit of 20!?
a) 2
b) 3
c) 5
d) 6
e) 8



Q2) What is the product of the factors of 432?
a) 2^90
b) 3^78
c) 2^40 * 3^30
d) 2^25 * 3^75
e) 2^5 * 3*10 * (2^25 * 3^30 - 1)



The easiest way to solve Q2 is as follows:

There is a formula to find the product of factors of a number 'n'.
=> (Prime factorization of n) ^ f/2 , where f is the total no. of factors.

Coming back to the quqestion, prime factorization of 432 = 2^4 * 3^3
Hence, total no.of factors = (4+1)*(3+1) = 5*4 = 20
The product of all factors of 432 = (2^4 * 3^3)^(20/2) = (2^4 * 3^3)^10 = 2^40 * 3^30 = C
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Re: PS - Rightmost nonzero digit & product of factors [#permalink] New post 08 May 2013, 05:15
holy wrote:
shrouded1 wrote:
mainhoon wrote:
What is the answer for 20!'s last digit?



Using the same approach I showed for 90!

Set 1 : {2,12}
Set 2 : {5,15}
Set 3 : {10,20}
Set 4 : {1,3,4,6,7,8,9, 11,13,14,16,17,18,19}

This reduces to (canceling 2s&5s in set 1 and set 2 + canceling 10s in set 3) :

Set 1 : {1,6} --> Last digit 6
Set 2 : {1,3} --> Last digit 3
Set 3 : {1,2} --> Last digit 2
Set 4 : {1,3,4,6,7,8,9, 11,13,14,16,17,18,19} --> Last digit 8*8=4

So overall last digit = 6*3*2*4 = 4

Hence last non-zero digit of 20! = 4



Can you please explain how you calculating the last digit for Set 4 ?


Set 4 is everything else {1,3,4,6,7,8,9,11,13,14,16,17,18,19, 21,23,24,26,27,28,29.........81,83,84,86,87,88,89}
notice the repeating pattern is set 4 - 1,3,4,6,7,8,9,1,3,4,6,7,8,9,........
and there are 9 such patterns (1,3,4,6,7,8,9) (11,13,14,16,17,18,19).....(81,83,84,86,87,88,89)

Multiply the digits in the patter 1*3*4*6*7*8*9 - the units digit is 8
Quote:
The "cyclicity" of 8 in a product is
8^1=8
8^2=64
8^3=512
8^4=4096
8^5=32768 and so on, the pattern is 8,4,2,6....


And as the pattern (1,3,4,6,7,8,9) repeats 9 times - so the units digit will be 8
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Re: PS - Rightmost nonzero digit & product of factors [#permalink] New post 09 May 2013, 04:48
shrouded1 wrote:
gurpreetsingh wrote:
Bump !!


There is a brute force way to solve this. The principle is the same, you have to remove all powers of 5 and a corresponding number of powers of 2 from the product. The product of the rest will give you the non-zero digit.

First of all what I'd do is split the numbers out into four sets. Set 1 is {2,12,22, ... ,82}. Set 2 is {5,15,25,...,85}. Set 3 is {10,20,30,...90} and Set 4 is everything else {1,3,4,6,7,8,9, 11,13,14,16,17,18,19, .... ,81,83,84,86,87,88,89}

Set 1,2,3 have 9 elements each and have all the powers of 5 within them. Set 4 has the remaining 7x9=63 elements

Lets process Set 3 first, I can take out all powers of 10, to get {1,2,3,4,5,6,7,8,9}
Then I take out the 5 and a corresponding 2 to get {1,1,3,4,1,6,7,8,9}

Now lets process Set 1 & 2, For every 5 I take out of set 2, I will take out a 2 from set 1 :
So set 1 becomes : {1,6,11,16,....,36,41} and set 2 : {1,3,5,7,9,11,13,15,17}

We are almost done now, the only powers of 5 left are in set 2 which I take out and against that I take out 4 in set 3 to leave me with with :

Set 1 : {1,6,.....,36,41} : Last digit is 6
Set 2 : {1,3,1,7,9,11,13,3,17} : Last digit is 7
Set 3 : {1,1,3,1,1,6,7,8,9} : Last digit 2
Set 4 : Which has a pattern to it and easy to show last digit is 8 (It has subsets ending in 1,3,4,6,7,8,9 each of which has last digit 8 and 9 such subsets. The "cyclicity" of 8 in a product is 8,4,2,6,repeat so with 9 subsets the answer will be 8)

Multiplying those out, the last digit of 90! is 2




The approach is brute force, but once you get a hang of it, it doesnt take very long to calculate it out. I would be very surprised if this is a GMAT level question, unless of course there is a trick to doing this which has completely evaded me :lol:


Hi,

I dont understand set 3.. Its ok you took out 2 and 5 , so that they become 1 and 1 respectively. But why 4 in set 3 becomes 1 ..Will you please explain?
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Re: PS - Rightmost nonzero digit & product of factors [#permalink] New post 09 May 2013, 07:02
bhanushalinikhil wrote:
GMAT TIGER wrote:
sergbov123 wrote:
Q2) What is the product of the factors of 432?

a) 2^90
b) 3^78
c) 2^40 * 3^30
d) 2^25 * 3^75
e) 2^5 * 3*10 * (2^25 * 3^30 - 1)


Thats a good & tough question:

a) 2^90 ......... cannot be because it doesnot have any multiple/power of 3.
b) 3^78......... cannot be because it doesnot have any multiple/power of 2.
c) 2^40 * 3^30......... can be because it has multiples/power of 2 & 3.
d) 2^25 * 3^75 ......... can be because it has multiples/power of 2 & 3.
e) 2^5 * 3*10 * (2^25 * 3^30 - 1) probably not.

Lets work on c and d.
432 = (2^4) (3^3)
Factor of 432 = 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 36, 48, 54, 72, 108, 144, 216 and 432.

Product of 432 = 2x3x4x6x8x9x12x16x18x24x27x36x48x54x72x108x144x216x432
= 2x3x 2^2 x 2x3 x 2^3 x 3^3 x 2^2x3 x 2^4 x 2x3^2 x 2^3x3 x 3^3 x 2^2x3^2 x 2^4x3 x 2x3^3 x 2^3x3^2 x 2^2x3^3 x 2^4x3^2 x 2^3x3^3 x 2^4x3^3
= 2^41 x 3^31

Hmmm ........ I got 2^41 x 3^31 instead of 2^40 x 3^30.

Answer should be C but how it is 2^40 x 3^30? I must have made a mistake somewhere....

Can you find it for me?
Thanx....


2x
3x
2^2 x
2x3 x
2^3 x
3^3(p) x
2^2x3 x
2^4 x
2x3^2 x
2^3x3 x
3^3 x
2^2x3^2 x
2^4x3 x
2x3^3 x
2^3x3^2 x
2^2x3^3 x
2^4x3^2 x
2^3x3^3 x
2^4x3^3

3^3(p) x For 9, you have written 3^3 instead of 3^2. But thts all I could find out. :?



I counted 40 2s, so that is ok...you miscounted...
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Re: What is the rightmost non-zero digit of 20!? [#permalink] New post 10 May 2013, 16:47
What is the answer to "last non-zero digit of 20!"

I want to verify that I'm doing it correctly...

I got 4...
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Re: What is the rightmost non-zero digit of 20!? [#permalink] New post 27 Jan 2014, 09:06
Question 1:

For the first digit after the 0s, we just multiply the unit digits excluding 0,2,5s
1*3*4*6*7*8*9 = 36288 (if you multiply step by step, drop the whole number and only use the unit digits of your solutions!)

This needs to be done twice, because the unit digits appear between 0 and 10 as well as between 10 and 20.

So you multiply the unit digit from 36288 with 36288: 8*8 = 64 --> First digit after the 0s is 4.

If you want to compute this with 90!, you've got to calculate 8^9 (or step by step with the unit digit of your solutions

64^4*8 -> 4^4 * 8 -> 6^2 * 8 -> 6*8 = 48 Thus the first digits (after the 21) 0s is 8. (back-checked with wolframalpha)

Question 2: (did not use timer, but guess between 1min and 1:30)

432 = 2^4 * 3^3

Attachment:
Screen Shot 2014-01-27 at 18.03.33.png
Screen Shot 2014-01-27 at 18.03.33.png [ 11.21 KiB | Viewed 153 times ]


According to the table, the solution is 2^40 * 3^30
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Re: What is the rightmost non-zero digit of 20!?   [#permalink] 27 Jan 2014, 09:06
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